Advertisements
Advertisements
प्रश्न
The points (2, -1), (-1, 4) and (-2, 2) are midpoints of the sides ofa triangle. Find its vertices.
उत्तर
Let A(x1,y1), B(x2, y2 ) and C(x3, y3) be the coordinates of the vertices of Δ ABC.
Midpoint of AB, i.e. D
D (2 , 1) = D `(("x"_1 + "x"_2)/2 , ("y"_1 + "y"_2)/2)`
`2 = ("x"_1 + "x"_2)/2 , ("y"_1 + "y"_2)/2` = -1
X1 + X2 = 4 .....(1) Y1 + Y2 = -2 ........(2)
Similarly,
X1 + X3 = -2 .....(3) y1 + y3 = 8 . ....(4)
X2 + X3 = -4 .....(5) Y2 + Y3 = 4 .....(6)
Adding (1), (3) and (5)
2(x1 + x2 + x3) = -2
x1 + x2 + x3 = -1
4 + x3 = -1 ... (from (1))
x3 = -5
From (3)
x1 - 5 = -2
x1 = 3
From (5)
x2 = -5 = -4
x2 = 1
Adding (2),(4) and (6)
2( Y1 + Y2 + Y3)= 10
Y1 +Y2 +Y3 = 5
-2 + Y3 = 5 [from(2)]
y3 = 7
From (4)
y1 + 7 = 8
y1 = 1
From (6)
y2 + 7 = 4
y2 = -3
The coordinates of the vertices of A.ABC are (3, 1), ( 1,-3) and ( -5,7)
APPEARS IN
संबंधित प्रश्न
ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find
1) Coordinates of A
2) An equation of diagonal BD
M is the mid-point of the line segment joining the points A(–3, 7) and B(9, –1). Find the coordinates of point M. Further, if R(2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q.
In the following example find the co-ordinate of point A which divides segment PQ in the ratio a : b.
P(2, 6), Q(–4, 1), a : b = 1 : 2
Point P is the midpoint of seg AB. If co-ordinates of A and B are (-4, 2) and (6, 2) respectively then find the co-ordinates of point P.
(A) (-1,2) (B) (1,2) (C) (1,-2) (D) (-1,-2)
The midpoints of three sides of a triangle are (1, 2), (2, -3) and (3, 4). Find the centroid of the triangle.
AB is a diameter of a circle with centre 0. If the ooordinates of A and 0 are ( 1, 4) and (3, 6 ). Find the ooordinates of B and the length of the diameter.
The centre ‘O’ of a circle has the coordinates (4, 5) and one point on the circumference is (8, 10). Find the coordinates of the other end of the diameter of the circle through this point.
The mid-point of the sides of a triangle are (2, 4), (−2, 3) and (5, 2). Find the coordinates of the vertices of the triangle
If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (−3, 2), then the coordinate of the other end of the diameter is
Point P is the centre of the circle and AB is a diameter. Find the coordinates of points B if coordinates of point A and P are (2, – 3) and (– 2, 0) respectively.
Given: A`square` and P`square`. Let B (x, y)
The centre of the circle is the midpoint of the diameter.
∴ Mid point formula,
`square = (square + x)/square`
⇒ `square = square` + x
⇒ x = `square - square`
⇒ x = – 6
and `square = (square + y)/2`
⇒ `square` + y = 0
⇒ y = 3
Hence coordinates of B is (– 6, 3).